Question

A process designed to remove organic sulphur from coal prior to combustion involves the reaction.
$X-S-Y+2NaOH\to X-O-Y+N{a}_{2}S+H_{2}O$
$CaC{O}_3\to CaO+C{O}_2$
$N{a}_{2}S+C{O}_2+H_{2}O\to N{a}_{2}C{O}_3+H_{2}S$
$CaO+H_{2}O\to Ca(OH{)}_2$
$N{a}_{2}C{O}_3+Ca(OH{)}_2\to CaC{O}_3+2NaOH$
In the processing of 320 metric tons of a coal having 1.0% sulphur content, how much limestone $\left(CaC{O}_3\right)$must be edecomposed to provide enough $Ca(OH{)}_2$ to regenerate the NaOH used in the original leaching step?

• A. 2.0 metric ton
• B. 4.0 metric ton
• C. 8.0 metric ton
• D. 10.0 metric ton

Answer 1

The correct option is D 10.0 metric ton

Weight of $CaC{O}_3=(320\times {10}^{6}gcoal)\left(\frac{1.0gS}{100gcoal}\right)\left(\frac{1molS}{32gS}\right)\left(\frac{1molCaC{O}_3}{molS}\right)\left(\frac{100gCaC{O}_3}{molCaC{O}_3}\right)$
$=\frac{320\times {10}^{6}1\times 1\times 1\times 100}{100\times 32}$
$=10\times {10}^{6}gCaC{O}_3$
$=10$ metric tone $CaC{O}_3$.
Hence, 10 metric tons of limestone $CaC{O}_3$ must be decomposed to provide enough Ca(OH)2 to regenerate the NaOH used in the original leaching step.Note: 1 metric ton $={10}^6$ g.